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yaroslaw [1]
3 years ago
8

Passengers in a high-flying jumbo jet feel their normal weight in flight, while passengers in an orbiting space vehicle do not,

because passengers in the space vehicle are
A) Without support forces.
B) Beyond the main pull of Earth's gravity
C) above Earth's atmosphere
D) all of the above
E) None of the above.
Physics
1 answer:
daser333 [38]3 years ago
4 0

Answer:E none of the above

Explanation:

When a space shuttle orbiting around the earth then earth gravitational force keeps the shuttle in its orbit by providing centripetal force. Tangential velocity makes the shuttle to revolve around the earth. Earth gravitational pull is significant to make the shuttle move in orbit while the Passengers feel weightlessness because they are falling downward but never able to reach as they move fast enough such that earth curves away them.                        

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You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

= 2.64 m/s

Explanation:

Working with the forces acting on the water in the pail at any point.

The weight of water is always directed downwards.

The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

Net force = W

Net force = centripetal force = (mv²/r)

W = mg

(mv²/r) = mg

r = 0.710 m

g = 9.8 m/s²

v² = gr = 9.8 × 0.71 = 6.958

v = √(6.958) = 2.64 m/s

Hope this Helps!!!

7 0
2 years ago
What is a lunar month
Ira Lisetskai [31]

Answer:

It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).

7 0
2 years ago
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When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f
elena-s [515]

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

3 0
3 years ago
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
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What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?
mote1985 [20]
The last choice. Two arrows and the arrow up is shorter than the arrow down. Since the guy is falling and he’s opened his chute, he’s slowing down but he’s still falling meaning the force of gravity is stronger than the air resistance.
8 0
3 years ago
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